Composition and Resolution of Forces

# Centre of Mass And Centre of Gravity

## Centre of Mass

A point on a body where the entire mass of the body can be assumed concentrated is known as centre of mass of the body.
A body consists of large number of particles. If an external force is applied directly on the centre of mass of the body, then a same motion will result in the entire body.

​Consider a rigid body consisting of large number of particles of masses $m_1, m_2, m_3, …, m_n$. Let $x_1, x_2, x_3, …, x_n$ be the respective distances of the particles from the axis $OY$. Let $y_1, y_2, y_3, …, y_n$ be the respective distances of the particles from the axis $OX$.

Then, the distance $x$ of the centre of mass of the body from $OY$ is given by, $x=\frac{m_1x_1+m_2x_2+m_3x_3+…+m_nx_n}{m_1+m_2+m_3+…+m_n}$

And, the distance $y$ of the centre of mass of the body from OX is given by, $y=\frac{m_1y_1+m_2y_2+m_3y_3+…+m_ny_n}{m_1+m_2+m_3+…+m_n}$

## Centre of Gravity

A rigid body consists of a large number of particles. Gravity of the earth acts on each individual particle. So, we can say that a large number of parallel forces are acting on the body which are directed vertically downward. Since all the forces are parallel, the net force on the body due to the gravity of the earth can be obtained by adding all the individual forces algebraically. The resultant of the forces on the body passes through a point known as centre of gravity.

​The resultant downward force, $W=m_1g+m_2g+m_3g+…+m_ng$ $W=(m_1+m_2+m_3+…+m_n)g$ $W=Mg$

where, $W$ is weight of the body, $M$ is mass of the body, $g$ is acceleration due to gravity and $(m_1, m_2, m_3, …, m_n)$ are the particles of the body.

The fixed point where the resultant downward force acts due to the gravity is known as centre of gravity (C.G.) of the body.

Consider a thin straight rod having particles’ weights $w_1, w_2, …$ lying along a straight line at distances $x_1, x_2, …$ from a fixed point $O$ on the line.

∴The fixed resultant or the weight of the body is, $w=w_1+w_2+…=\sum w_i$

And the point of application of the weight is at a distance,

$\bar{x}=\frac{w_1x_1+w_2x_2+…}{w_1+w_2+…}$$=\frac{\sum w_ix_i}{\sum w_i} \text{ from O}$

This point is a fixed point on the rod and is called the centre of gravity of the rod.

If the rigid body is a plane, then the weight of the body acts at a point $(\bar{x},\bar{y})$ such that,

$\bar{x}=\frac{\sum w_ix_i}{\sum w_i} \;\;\; \text{and,} \;\;\; \bar{y}=\frac{\sum w_iy_i}{\sum w_i}$

This point $(\bar{x},\bar{y})$ is a unique point which denotes the position of any one of the particles and is the centre of gravity of the body.

If a body is uniform and homogeneous having standard geometric shape, then its centre of gravity generally lies at its geometric centre.